Question

The string shown in the figure is passing over a small smooth pulley rigidly attached to trolley A. If the speed of the trolley is constant and equal to $v_A$, speed and acceleration of block B at the instant is $v_B$ and $a_B$ respectively, which of the following option(s) is/are correct?

• A. $v_B=v_A,a_B=0$
• B. $a_B=0$
• C. $v_B=\frac{3}{5}{v}_A$
• D. $a_B=\frac{16v_A^2}{125}$

Answer 1

The correct option is D $a_B=\frac{16v_A^2}{125}$



From the figure,
$l_1+l_2=\text{constant}$
(segments of same string)
$\Rightarrow (y-h)+\sqrt{x^2+h^2}=\text{constant}$
Differentiating w.r.t time $t$, we get
$\frac{dy}{dt}+\frac{x}{\sqrt{x^2+h^2}}\frac{dx}{dt}=0$
$\Rightarrow \frac{dy}{dt}=-\frac{x}{\sqrt{x^2+h^2}}\frac{dx}{dt}$
At the instant shown in figure,
$\frac{x}{\sqrt{x^2+h^2}}=\frac{3}{\sqrt{3^2+4^2}}=\frac{3}{5}$
$\frac{dx}{dt}=v_A$
$\frac{dy}{dt}=-\frac{3}{5}{v}_A$
Since,
$v_B=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}>\frac{dy}{dt}$
$\Rightarrow |v_B|>\frac{3}{5}{v}_A$
We will now differentiate $\frac{dy}{dt}$ to obtain acceleration of B.
$\frac{dy}{dt}=-\frac{x}{\sqrt{x^2+h^2}}.v_A$
$\Rightarrow \frac{d^{2}y}{d{t}^2}=-\frac{(x^2+h^2{)}^{1/2}.1.\frac{dx}{dt}-x.\frac{1}{2}(x^2+h^2{)}^{-1/2}.2x.\frac{dx}{dt}}{(x^2+h^2)}.v_A$
( $v_A$ is constant)
$\frac{dy}{dt}=\frac{h^2}{(x^2+h^2{)}^{3/2}}.\frac{dx}{dt}.v_A$
$=v_A^2\frac{h^2}{(x^2+h^2{)}^{3/2}}$
$\therefore$ at the instant shown in figure,
$a_B=v_A^2\frac{4^2}{(3^2+4^2{)}^{3/2}}=v_A^2\frac{4^2}{(5{)}^3}=\frac{16}{125}{v}_A^2$